Will PbCl2 precipitate on mixing 1 ml of 3  10-3 M NaCl solution with 2 ml of 1  10-3 M Pb(NO3)2 solution?

To answer this question, we need to calculate the reaction quotient (Q) and compare it with the solubility product constant (Ksp) for PbCl2.

The solubility product constant, Ksp, for PbCl2 is 1.7 x 10^-5 at room temperature.

First, we need to calculate the final concentrations of Cl- and Pb2+ ions in the solution.

The total volume of the solution is 1 ml + 2 ml = 3 ml.

The final concentration of Cl- ions = initial concentration x initial volume / final volume = 3 x 10^-3 M x 1 ml / 3 ml = 1 x 10^-3 M.

The final concentration of Pb2+ ions = initial concentration x initial volume / final volume = 1 x 10^-3 M x 2 ml / 3 ml = 0.67 x 10^-3 M.

The reaction quotient, Q, is given by [Pb2+][Cl-]^2.

Substituting the calculated concentrations into this expression gives Q = (0.67 x 10^-3) x (1 x 10^-3)^2 = 0.67 x 10^-9.

Since Q < Ksp, no precipitate will form. Therefore, PbCl2 will not precipitate when 1 ml of 3 x 10^-3 M NaCl solution is mixed with 2 ml of 1 x 10^-3 M Pb(NO3)2 solution.